LeetCode_Word Ladder

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

Only one letter can be changed at a time
Each intermediate word must exist in the dictionary
For example,

Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.

　　DFS 小数据ＡC：

class Solution {
public:
   bool check(const string & a, const string &b)
    {
        int num = 0;
        if(a.size() != b.size()) return false;
        for(int i = 0; i< a.size() ; i++)
        {
            if(a[i] != b[i])
               num++;
        }
        return num == 1;
    }
    void DFS(const string &start, const string &end, unordered_set<string> &dict, vector<bool> &flag, int nums){
    
            if(start == end ){
                res = res > nums ? nums : res;
                return ;
            }
            int i;auto it = dict.begin();
           for( i= 0; it != dict.end(); it++,i++) 
               if(flag[i] == false && check(start,*it))
               {
                    flag[i] = true;
                    DFS(*it,end,dict, flag, nums+1);
                    flag[i] = false;
               }
    
    }
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
          res = dict.size() + 1 ;
          vector<bool> flag(dict.size(), false);
          DFS(start, end, dict, flag, 0);
          if(res  ==  dict.size() + 1) return 0;
          return res +1 ;
    }
private : 
  int res;
};

 BFS：　过大数据

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(start.size() != end.size()) return 0;
        if(dict.size() == 0) return 0;
        
        queue<string> myqueue, myqueueT;
        myqueue.push(start);
        int depth = 1;
        
        while(!myqueue.empty()){
          depth++;
          while(!myqueue.empty()){
            string str = myqueue.front();
            myqueue.pop();
            for(int i = 0; i < str.size() ; i++){
                char temp = str[i] ;
                for(char c = 'a'; c <= 'z' ;c++){
                     if(c == temp) continue;
                     str[i] = c;
                     if(str == end) return depth;
                     auto it = dict.find(str) ;
                     if(it != dict.end() ){
                        myqueueT.push(str);
                        dict.erase(it);
                     }
                }
                str[i] = temp;
            }
          }
          myqueue.swap( myqueueT);
        }
        //don't find 
        return 0;
    }
};